1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 655

lzl > R, has the form

+oo

f( z ) = L B kZ k = ···+ B-n - +"·+-+ B-1 B o+B1z+···+Bnz n +···

zn z

k=-oo

(9.2-2)

Hence we have

( )

+oo

1 -k Bn B1

g(z) = f z - = '"" L.J Bkz = .. · + - + zn .. · + -z

k=-oo

+Bo + B-1Z + · · · + B-nZn + · · · (9.2-3)

this expansion being valid in any circular ring with center at the origin and

contained in 0 < lzl < 1/ R.

It follows from (9.2-3) that z = 0 is a pole of order m of g if En = 0 for
n > m but Em -:f 0, and that z = 0 is an isolated essential singularity of g
if En -:f 0 for infinitely many values of n. As a consequence, z = oo is a pole

of order m of f if only a finite number of powers with positive exponents

appear in (9.2-2), the highest being Bmzm, and z = oo is an essential

singularity of f if an infinite number of powers with positive exponents

occur in (9.2-2). Letting /3 = sup {k: Bk -:f O} we may say .that f has a

pole at oo if 0 < /3 < +oo, and an essential singularity if /3 = +oo. If

/3 :::; 0 then f is regular at oo.

Examples 1. Since

eZ 1 1 1 1 1 Z

f(z) = z3 = z3 + z2 + 2! ; + 3! + 4! + .. ,

valid for 0 < lzl < oo, the function ez /z^3 has a pole of order 3 at z = 0.

2. Let f(z) = 1 - 4z + z^2. Since there is a finite number of terms
   containing powers of z with positive exponents, and the highest power is
   z^2 , this function has a double pole at oo.


3. f(z) = sinz = z - z^3 /3! + z^5 /5! - · .. has an isolated essential
   singularity at oo.

or

From (9.2-1) it follows that

. 1
f(z)= ( z-am ) [A-m+A-m+1(z-a)+ .. ·+Ao(z-ar+ ...

f(z) = h(z)

(z - a)m

(9.2-4)

where h(z) = A_m + A-m+ 1 (z - a)+··· is analytic at z =a and h(a) =
A-m -:f 0.